Keeping the Fibonacci rule of adding the latest two to get the next but starting from 2 and 1 (in this order) instead of 0 and 1 for the (ordinary Fibonacci numbers) one defines the series of Lucas Numbers:
L(n) = L(n-1) + L(n-2) for n>1
L(0) = 2
L(1) = 1
Here are some more values of L(n) together with the Fibonacci numbers
for comparison:
F(n): 0 1 1 2 3 5 8 13 21 34 55 ...
L(n): 2 1 3 4 7 11 18 29 47 76 123 ...
With the above information in background prove the following formulas:
a. F(n-1)+ F(n+1) = L(n)
b. L(n-1) + L(n+1)= 5*F(n)
c. F(n)* L(n) = F(2n)
Part c) can be proven two ways:
1) The method of induction (I have not gotten this to work, and I wish I had!)
2) Binet's formulae
We go for proof 2.
We express F(n) and L(n) using Binet's formulae:
Note: the formulae are quite approximate, and very approximate for small n. E.g.: they are not valid for n < 4
P is for Phi (the Golden Ratio):
P= {1+sqrt(5)}/2 = 1.618...
and p is for phi, Phi's inverse:
p=2/{1+sqrt(5)} = P-1 = 0.618...
Here are Binet's Formulae:
F(n) = [P^n - (-p)^n)] / sqrt(5)
L(n) = [P^n + (-p)^n]
So,
F(n) L(n) = {[P^n - (-p)^n)] / sqrt(5)} [P^n + (-p)^n]
= [P^(2n) - (-p)^(2n)] / sqrt(5)
= F(2n)
Q.E.D.
Edited on September 9, 2018, 11:52 pm
part c) soln. , using Binet' formulae
